b Use the formula y=a(1+ rn)^(nt) where n is the number of times the initial value is compounded in a year.
C
c Use the formula y=a(1+ rn)^(nt) where n is the number of times the initial value is compounded in a year.
D
d Use the formula y=ae^(rt).
A
a About 39.52 years.
B
b About 38.66 years.
C
c About 38.38 years.
D
d About 38.38 years.
Practice makes perfect
a The balance, y, can be modeled by an Exponential-Growth-Model.
y=a(1+r)^t
In this function, a is the initial amount, r is the percent increase expressed in decimal form, and t is the time in years. We know that the account starts with a= 100 and pays 6 % annual interest which, expressed in decimal form, can be written as r= 0.06.
y= 100(1+ 0.06)^t ⇔ y=100(1.06)^t
By setting y equal to 1000, we can solve for t.
To solve this equation, we need to use logarithms. Notice that the base of the power on the left-hand side is 1.06. Does this mean we have to use a logarithm with a base of 1.06? No, as long as the same logarithm is applied to both sides of the equation, we can use any logarithm we want.
b If the frequency of compounding is quarterly, we have to use a formula that describes compound interest. Like in Part A, we already know that a= 100 and r= 0.06.
y=a(1+r/n)^(nt) ⇒ y= 100(1+0.06/n)^(nt)
In the formula, n is the number of times the initial principal is compounded per year. Since there are 4 quarters in a year, we have n=4
y=100(1+0.06/4)^(4t)
Like in Part A, we will set y equal to 1000 and solve for t.
When compounded daily, it will take about 38.38 years for the principal to reach $1000.
d When interest is compounded continuously, we need to use a different formula for compounded interest.
y=ae^(rt)
Like in previous parts, a is the initial value and r is the annual interest rate expressed as a decimal.
y=100e^(0.06t)
By substituting y=1000, we can solve for t.
